We throw 2 fair dice together 6 times. What is the probability of getting 3 "doubles", i.e. 3 times the same number in both dice?

My attempt: For the first draw of 2 dice together, the sample space has 36 possible outcomes and the ones with the same number in both dice are $(1,1), (2,2)...(6,6)$ so a total of 6. The probability for one throw is $P_1 = \frac {6}{36} = \frac {1}{6}$

Therefore for 2 such throws, the probability to get doubles is $(\frac {1}{6})^2$.

The probability to get doubles in 3 out of 6 throws is

$(\frac {1}{6})^3*(\frac {5}{6})^3$ ?? Is this correct?

or maybe $\frac {6^3}{36^6}$?

Thank you!

  • $\begingroup$ For example, probability of getting one double for two throws is $2(1/6)(5/6)$. $\endgroup$
    – Atbey
    Commented Jan 14, 2021 at 9:48

1 Answer 1


As you observed, the probability of obtaining doubles on any one throw of a pair of dice is $1/6$ since $6$ of the $36$ possible outcomes result in doubles. Since the probability of obtaining doubles is the same for each of the six trials, this is a binomial distribution problem.

The probability of obtaining exactly $k$ successes in $n$ trials is $$\Pr(X = k) = \binom{n}{k}p^k(1 - p)^{n - k}$$ where $p^k$ is the probability of $k$ successes, $(1 - p)^{n - k}$ is the probability of $n - k$ failures, and $\binom{n}{k}$ counts the number of ways in which exactly $k$ of the $n$ trials could result in a success.

Here, we will define a "success" as a double, so $p = 1/6$. Since there are six trials, $n = 6$. Hence, the probability to get exactly three doubles in six trials is $$\Pr(X = 3) = \binom{6}{3}\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^3$$ In your attempt, you did not take into account the number of ways exactly three of the rolls could result in a success.

If you were instead interested in the probability of at least three doubles in six trials, you would add the probabilities of three or more successes. $$\Pr(X \geq 3) = \sum_{k = 3}^{6} \binom{6}{k}\left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{6 - k}$$

  • $\begingroup$ Thank you, I hadn't taken into account the term $\binom{6}{3}$. $\endgroup$ Commented Jan 14, 2021 at 10:46

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